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Section 2.6: Probability and
Expectation
Practice HW (not to hand in)
From Barr Text
p. 130 # 1, 2, 4-12
• Cryptanalyzing the Vigene re cipher is not
trivial process. The purpose of this section and
the next section is to show a probabilistic
method that allows one to determine the likely
keyword length which is the first step in
breaking this cipher. In this section, we review
the basics of counting and probability.
Permutations
• Permutations represent an ordered listing
of a set. Before defining what a
permutation is, we give an important
preliminary definition.
Definition
• Factorial. If n is a positive integer, then n
factorial, denoted as n!, is defined to be
n !  n( n  1)( n  2) 2  1
Note: 1! 1, 0! 1
Example 1: Calculate 3!, 5!, and 10! .
9!
Solution:
Example 2: John, Mary, and Sue have bought
tickets together for a basketball games. In how
many ways could they arrange themselves in the
3 seats? In 6 seats?
Solution:
Formal Definition of a Permutation
• A permutation of a set of objects is a listing of
the objects in some specified order.
• Our goal in this section will be to count the
number of permutations for specific objects.
This next three examples illustrate how this can
be done.
Example 3: A baseball team is made up of 9
players and the manager wants to construct
batting orders.
a. How many total possible batting orders can
the manager construct?
Solution:
b. How many batting orders can the manager
construct if the pitcher must bat last?
Solution:
c. How many batting orders can the manager
construct if the shortstop and pitcher bats
eighth or ninth?
Solution:
Example 4: If there are 50 contestants in a
beauty pageant, in how many ways can the
judges award first, second, and third prizes?
Solution:
Example 5: How many license plates can be
made if each plate consists of
a. two letters followed by three digits and
repetition of letters and digits is allowed?
Solution:
b. two letters followed by three digits and no
repetition of letters or digits is allowed?
Solution:
c. if the first digit cannot be a zero and no
repetition of letters or digits is allowed?
Solution:
Note
• Given a collection of r objects, the number of
ordered arrangements (permutations) of r
objects taken from n objects, denoted as P(n, r)
is given by
n!
P(n, r ) 
(n  r ) !
(1)
Example 6: Find
Solution:
P(10,4) .
• Equation (1) can be used to count the number
of permutations as the next example indicates.
Example 7: Going back to Example 4, find the
number of ways that judges award first, second,
and third prizes in a beauty contest with 50
contestants using equation (1).
Solution:
Combinations
We now want to consider how we can count
different arrangements when the order of the
arrangements does not matter.
Example 8: Suppose you have five clean shirts
and are going to pack two for a trip. How many
ways can you select the two shirts? Compare the
difference in this problem when the order the
shirts are packed matters and when it does not.
Solution:
Definition of Combinations
• A combinations is an unordered set of r objects
chosen from a set of n is called a combination
of r objects chosen from n objects. The number
of different combinations of r unordered objects
chosen from n unordered objects is
n!
C ( n, r ) 
r ! (n  r ) !
(2)
Example 9: Compute C (7,4) and C (10,3) .
Solution
Example 10: Referring back to Example 8, use
equation (2) to find the number of ways you can
choose 2 shirts from 5 total to go on a trip.
Solution:
• A permutation is a listing of objects where the
order of the objects in the list is important.
Usually, some ranking or order of the list is
given to note its importance. In a combination,
the order of the objects in the list is not
important. Thus, counting the number of
permutations and combinations is different as a
result. The following examples illustrate the
difference between the two.
Example 11: Suppose Radford’s Honors
Academy wants to select four students out of nine
total for a committee to go to an honors
convention. How many ways can the committee
of four be chosen?
Solution:
Example 12: Suppose Radford’s Honors
Academy wants to select four students out of nine
total for a committee to go to an honors
convention. For the four students selected, one
will serve as President, one as Vice President,
one as Secretary, and the other as Treasurer for
the committee. How many ways can this
committee be selected?
Solution:
Basic Probability
• We now discuss some basic concepts of
probability. We start out with a basic definition.
• Definition: The sample space of an experiment
is the set of all possible outcomes of an
experiment.
Example 13: Determine the sample space of the
single toss of a die.
Solution:
Definition
• An event is any subset of the sample space.
Example 14: List some events for sample space
consisting of a single roll of a die.
Solution:
Definition of Probability
• The probability of and event is a number
between 0 and 1 that represents the chance of
an event occurring. If A is an event , then
Probabilit y that
the event A occurs
 P( A) 
the number of ways that the event A can occur
total number of outcomes that occurs in the sample space
Example 15: On a single toss of a die, find the
probability of rolling
a. a 5.
Solution:
b. an even number.
Solution:
c. the number showing is no less than a 5.
Solution:
d. roll that is not a 2.
Solution:
e. a 7.
Solution:
Facts about Probability
Given the probability P of an event occurring.
1. 0  P  1
2. Given two events A and B that are mutually
exclusive (both events A and B are separate,
they can’t happen at the same time), then
P( A or B)  P( A)  P( B)
Example 16: For the single die roll example,
explain why rolling a 4 and a 6 are mutually
exclusive events. Then find the probability of
rolling a 4 or a 6.
Solution:
3. Given the probability of an event A,
P(not A)  1  P( A)
Example 17: For the single die roll example,
use fact 3 to determine the probability of not
rolling a 5.
Solution:
4. The sum of all the probabilities of mutually
exclusive events in a sample space is equal to 1.
Example 18: Find the probability on a single die
roll of rolling a 1, 2, 3, 4, 5, 6.
Solution:
Example 19: Suppose you toss two ordinary die
and observe the sum of roll.
a. What is the sample space of the event?
Solution:
b. What is the probability that the sum of the
numbers of the die is a 7? A 5?
Solution:
c. Find the probability that the sum of the
numbers is at most 5.
Solution:
d. Find the probability that exactly one of the
numbers is a 5.
Solution:
Probability of Simultaneous Events
Suppose we have two events that can occur
simultaneously, that is, can be done
independently of one another. Then we can find
the probability of both events occurring by using
the following multiplication principle of probability.
Multiplication Principle of Probability
If two (ordered or labeled) experiments A and B
can be conducted independently, that is both can
be done simultaneously, then if P ( A) and
P(B) represents the probabilities of these
separate events occurring, then the probability in
the compound experiment of both outcomes
occurring is
P( A and B)  P( A)  P( B).
Example 20: A pot contains alphabet letters
consisting of 300 A’s, 154 B’s, 246 C’s, and 500
D’s. Suppose we draw two letters from the pot
without replacement. Find the probability that
a. both letters are C’s.
Solution:
b. both letters are D’s.
Solution:
c. The first letter is a A and the second is a B.
Solution:
d. The two letters are A and B.
Solution:
e. Neither letter is a C.
Solution:
f. The letters are identical.
Solution: